Appendix: Matlab Examples

GUIDE: Mathematics of the Discrete Fourier Transform (DFT). Appendix: Matlab Examples

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Appendix: Matlab Examples

Here’s how Fig. 6.1 was generated in Matlab:

>> x = [2 3];                  % coordinates of x
>> origin = [0 0];             % coordinates of the origin
>> xcoords = [origin(1) x(1)]; % plot() expects coordinate lists, not endpoints
>> ycoords = [origin(2) x(2)];
>> plot(xcoords,ycoords);      % Draw a line from origin to x
Mathematica can plot a list of ordered pairs:
In[1]:
    ListPlot[{{0,0},{2,3}},PlotJoined->True]; (* Draw a line from (0,0) to (2,3) *)

In Matlab, the mean of the row-vector $x$ can be computed as

\

or by using the built-in function mean().

In Matlab, if x = [x1 … xN] is a row vector, we can compute thetotal energy as

\

Matlab has a function orth() which will compute an orthonormal basis for a space given any set of vectors which span the space.

>> help orth

ORTH Orthogonalization. Q = orth(A) is an orthonormal basis for the range of A. Q’*Q = I, the columns of Q span the same space as the columns of A and the number of columns of Q is the rank of A.

    See also QR, NULL.</pre>Below is an example of using <tt>orth()</tt> to orthonormalize a <a href="http://mathworld.wolfram.com/linearlyindependent.php">linearly

independent basis set for $N=3$:

% Demonstration of the Matlab function orth() for
% taking a set of vectors and returning an orthonormal set
% which span the same space.
v1 = [1; 2; 3];  % our first basis vector (a column vector)
v2 = [1; -2; 3]; % a second, linearly independent column vector
v1’ * v2         % show that v1 is not orthogonal to v2

ans =

 6

V = [v1,v2] % Each column of V is one of our vectors

V =

 1     1
 2    -2
 3     3

W = orth(V) % Find an orthonormal basis for the same space

W =

0.2673    0.1690
0.5345   -0.8452
0.8018    0.5071

w1 = W(:,1) % Break out the returned vectors

w1 =

0.2673
0.5345
0.8018

w2 = W(:,2)

w2 =

0.1690

-0.8452 0.5071

w1’ * w2 % Check that w1 is orthogonal to w2 (to working precision)

ans =

2.5723e-17

w1’ * w1 % Also check that the new vectors are unit length in 3D

ans =

 1

w2’ * w2

ans =

 1

W’ * W % faster way to do the above checks (matrix multiplication)

ans =

1.0000    0.0000
0.0000    1.0000

% Construct some vector x in the space spanned by v1 and v2: x = 2 * v1 - 3 * v2

x =

-1
10
-3

% Show that x is also some linear combination of w1 and w2: c1 = x’ * w1 % Coefficient of projection of x onto w1

c1 =

2.6726

c2 = x’ * w2 % Coefficient of projection of x onto w2

c2 =

-10.1419

xw = c1 * w1 + c2 * w2 % Can we make x using w1 and w2?

xw =

-1.0000 10.0000 -3.0000

error = x - xw

error =

1.0e-14 *

0.1332
     0
     0<a href="http://www-ccrma.stanford.edu/~jos/interpolation/lp_norms.php">norm</a>(error)       % typical way to summarize a vector error

ans =

1.3323e-15

% It works!

% Now, construct some vector x NOT in the space spanned by v1 and v2: y = [1; 0; 0]; % Almost anything we guess in 3D will work

% Try to express y as a linear combination of w1 and w2: c1 = y’ * w1; % Coefficient of projection of y onto w1 c2 = y’ * w2; % Coefficient of projection of y onto w2 yw = c1 * w1 + c2 * w2 % Can we make y using w1 and w2?

yw =

0.1000
0.0000
0.3000

yerror = y - yw

yerror =

0.9000
0.0000

-0.3000

norm(yerror)

ans =

0.9487

% While the error is not zero, it is the smallest possible % error in the least squares sense. % That is, yw is the optimal least-squares approximation % to y in the space spanned by v1 and v2 (w1 and w2). % In other words, norm(yerror) <= norm(y-yw2) for any other vector yw2 made % using a linear combination of v1 and v2.
% In yet other words, we obtain the optimal least squares approximation % of y (which lives in 3D) in some subspace W (a 2D subspace of 3D) % by projecting y orthogonally onto the subspace W to get yw as above. % % An important property of the optimal least-squares approximation % is that the approximation error is orthogonal to the the subspace % in which the approximation lies. Let’s show this:

W’ * yerror % must be zero to working precision

ans =

1.0e-16 *

-0.2574 -0.0119

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